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Proof step 6 from proof size-del-nodups (previous step)

#(a0 + x0) = m + 1, a ∈ a0 + x0, ∀ m, a. # x0 = m + 1 ∧ a ∈ x0 → #(x0 -l a) = m ∨ x0 = [] ∨ dups(x0) ⊦ #((a0 + x0) -l a) = m, a0 + x0 = [], dups(a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a ∈ x0, ∀ m0, a1. # x0 = m0 + 1 ∧ a1 ∈ x0 → #(x0 -l a1) = m0 ∨ x0 = [] ⊦ 1 + #(x0 -l a) = # x0, dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: del-y: ⊦ (a ' + x) -l a = x -l a;%in-03: ⊦ ¬ a ∈ x → x -l a = x;%del-n: ⊦ a ≠ b → (b ' + x) -l a = b ' + (x -l a);% ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x; (ws)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ' + x ≠ []; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)